Timeline for Alternating power series $\sum_{k\geq 0}(-1)^k z^{(2k+1)^2}$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2017 at 16:47 | comment | added | Christian Remling | @Bombyxmori: $a_n=0$ unless $n=(2j+1)^2$. | |
| Jun 13, 2017 at 2:06 | comment | added | Bombyx mori | Hey, can I ask how $|a_{n_j-k}|\rightarrow 0$ in our case? I thought $a_{i}=(-1)^{i}$. Not sure what I am missing at here... | |
| Jun 10, 2017 at 9:05 | comment | added | მამუკა ჯიბლაძე | and similarly to what Zagier does in his paper "Vassiliev invariants and a strange identity related to the Dedekind eta-function", I believe one can prove that these radial limits coincide with the result of substituting roots of unity in the above series (on roots of unity this actually terminates, giving infinity on "odd-odd" roots of unity and a well-defined explicit expression on all other roots of unity) | |
| Jun 10, 2017 at 9:02 | comment | added | მამუკა ჯიბლაძე | In fact, from the Coogan-Ono paper I linked in a comment to the question, one has$$g(z)=\frac{z}{1+z^2}+\frac{z^3 \left(1-z^2\right)}{\left(1+z^2\right) \left(1+z^6\right)}+\frac{z^5 \left(1-z^2\right) \left(1-z^6\right)}{\left(1+z^2\right) \left(1+z^6\right) \left(1+z^{10}\right)}+\frac{z^7 \left(1-z^2\right) \left(1-z^6\right) \left(1-z^{10}\right)}{\left(1+z^2\right) \left(1+z^6\right) \left(1+z^{10}\right) \left(1+z^{14}\right)}+...$$ | |
| Jun 10, 2017 at 8:42 | comment | added | მამუკა ჯიბლაძე | From numerical experiments, $g$ seems to have well-defined radial limits at all roots of unity. It tends to infinity very quickly along radii to $e^{2\pi i\frac pq}$ with both $p$ and $q$ odd, and to certain finite limit when only one of $p$, $q$ is even. | |
| Jun 10, 2017 at 3:12 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jun 10, 2017 at 1:36 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jun 10, 2017 at 1:28 | history | answered | Christian Remling | CC BY-SA 3.0 |